Results 1 to 5 of 5

Math Help - finding the circle from 3 points

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    8

    finding the circle from 3 points

    How am I supposed to approach this problem? the points are A(0, 0), B(6, 0) and C (4, 4)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    904
    Thanks
    27

    finding the circle from three points

    Quote Originally Posted by ziyadbasheer View Post
    How am I supposed to approach this problem? the points are A(0, 0), B(6, 0) and C (4, 4)
    Plot the three points and draw the triangle.assuming you want the circumscribed circle erect the perpendicular bisectors of two sides.use 0,0 and 6,0 for one of them.from slope diagrams find the slope of each line (side). then find the center of each line. write two equations in x and y for linesperpendiculartoeach(point slope formula) at the midpoints. Solve the two equations for x and y the center of the circle.finallydetermine the
    radiusby distance formular between center and a vertex.if this makes sense and you can proceed repost to write the circle equationor if you need more help


    bjh
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    8
    Thank you for replying, but I can't understand what you said. I don't know anything about math formulas
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    The general equation of a circle is

    (x - h)^2 + (y - k)^2 = r^2

    where r is the radius and (h, k) is the centre.


    You know that (0, 0), (6, 0), (4, 4) are three points on the circle. So you can get three equations in three unknowns:


    (0 - h)^2 + (0 - k)^2 = r^2

    (6 - h)^2 + (0 - k)^2 = r^2

    (4 - h)^2 + (4 - k)^2 = r^2.



    Equations 1 and 2 are equal, so

    (-h)^2 + (-k)^2 = (6 - h)^2 + (-k)^2

    (-h)^2 = (6 - h)^2

    h^2 = 36 - 12h + h^2

    0 = 36 - 12h

    12h = 36

    h = 3.


    Substituting into the second and third equations gives:

    (6 - 3)^2 + (0 - k)^2 = r^2 and (4 - 3)^2 + (4- k)^2 = r^2.


    Therefore 9 + k^2 = 1 + (4 - k)^2

    8 + k^2 = 16 - 8k + k^2

    8k = 8

    k = 1.



    Substituting into the third equation:

    (4 - 3)^2 + (4 - 1)^2 = r^2

    1 + 9 = r^2

    10 = r^2

    r = \sqrt{10}.



    So finally, the equation of the circle is

    (x - 3)^2 + (y - 1)^2 = (\sqrt{10})^2.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by ziyadbasheer View Post
    How am I supposed to approach this problem? the points are A(0, 0), B(6, 0) and C (4, 4)
    Hi ziyadbasheer,

    As (0,0) and (6,0) are two points on the circle,
    the circle's centre lies on the line that is the perpendicular bisector
    of the line segment starting at (0,0) and ending at (6,0).

    This line segment is a chord of the circle and lies on the x-axis.
    Hence the perpendicular bisector is vertical (parallel to the y-axis)
    and passes through (3,0) which is the midpoint of the line segment from (0,0) to (6,0).

    The circle centre also lies on the perpendicular bisector of the line segment
    (which is another chord of the circle) going from (0,0) to (4,4).

    The midpoint of this line segment is (2,2) and since the line has a 45 degree angle with the x-axis, going upwards from (0,0) to (4,4)
    then the perpendicular bisector goes downwards at 45 degrees from (2,2)
    and hence it crosses the x-axis at (4,0).

    For this line, as x increases by 1, y decreases by 1.

    Hence when it reaches x=3, (an increase of 1 in x) it will have decreased by 1
    in the y direction.

    Therefore the two perpendicular bisectors cross at (3,1).

    This is the circle centre.
    You can now draw the circle.

    It's radius can be calculated with Pythagoras' theorem if you like

    1^2+3^2=r^2=1+9=10

    r=\sqrt{10}
    Attached Thumbnails Attached Thumbnails finding the circle from 3 points-circle-3-points.jpg  
    Last edited by Archie Meade; April 24th 2010 at 06:55 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: August 19th 2011, 06:59 AM
  2. Replies: 16
    Last Post: June 10th 2011, 06:49 AM
  3. Replies: 4
    Last Post: May 2nd 2010, 03:34 PM
  4. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  5. Replies: 2
    Last Post: May 23rd 2007, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum